Question

How much Al is required to form alumina with 12 g of oxygen

Answer 1

the required equation for the reaction is

4Al+3O2=2Al2O3

thus (4×27)gram Al reacts with 96g of Oxygen so 1g of Oxygen reacts with ((4×27)/96)g of Al

thus 12g of Oxygen will require (((4×27)/96)×12)gram Al=(108/8) grams of Al=13.5grams of Al

Answer 2

Answer:

The amount of aluminium equals to 13.5g, is required to form alumina with 12g of oxygen.

Explanation:

The chemical reaction of formation of alumina:-

4Al   +   3O₂   $\longrightarrow$   2Al₂O₃

From the above reaction  we can say that three moles of oxygen will react with four moles of aluminium.

The atomic mass of  oxygen = 16g/mol

Given, the mass of oxygen = 12g

The number of moles of oxygen for given mass = 12/32 = 0.375 moles

3 moles of oxygen react with aluminium = 4 moles

0.75 moles of oxygen will react with aluminium $=\frac{4}{3} \times 0.375 = 0.5 mole$

The mass of Aluminium required = 0.5× 27 = 13.5g

Therefore, 13.5 g of aluminium is required to form alumina with 12g of oxygen.

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